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I was considering some number theory problems which inspired me to write the following conjecture, which bears some resemblance to the Catalan problem, but is in fact different:

Fix two distinct sequences of primes $p_{1}, ..., p_{n}$ and $q_{1}, ..., q_{m}$. Do there exist infinitely many sequences of naturals $a_{1}, ..., a_{n}$, $b_{1}, ..., b_{m}$ such that:

$p_{1}^{a_{1}} ... p_{n}^{a_{n}} - q_{1}^{b_{1}} ... q_{m}^{b_{m}} = 1$?

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With one factor each side, do you know what happens with $p=3$ and $q=2$? – Mark Bennet Jun 15 '12 at 20:30
It only has one solution in that case – Pedro Jun 15 '12 at 20:38
I'm interested in cases where at least one of $n, m$ is bigger than 2 – Pedro Jun 15 '12 at 20:51
up vote 2 down vote accepted

Thue proved ( that the equation $$A \cdot X^k - B \cdot Y^k = 1$$ (for fixed $A$, $B$, and $k$) has only finitely many integral solutions if $k \ge 3$.

Fix two sets of primes $p_i$ and $q_j$. Your equations give integral solutions to a finite number of Thue equations, and thus there can be at most finitely many solutions.

For example (to be very explicit about the construction), any solution to $2^a 3^b - 5^c 7^c = 1$ yields a solution to $$A x^3 - B y^3 = 1$$ with $A \in \{1,2,4,3,6,12,9,18,36\}$ and $B \in \{1,5,25,7,35,175,49,245,1225\}$.

More generally, this problem falls under the broader class of problems known as $S$-unit equations (, which are well studied.

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Conditional on the abc conjecture, the answer is no. We have $1 + \prod q_i^{b_i} = \prod p_i^{a_i}$ and the abc conjecture implies that $$\prod p_i^{a_i} \le (\prod p_i \prod q_i)^2$$

for all but finitely many choices of the $a_i$ and $b_i$. But this condition itself can only hold for finitely many choices of the $a_i$.

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